For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 e-lambda t). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging exponential profile starting from 1, which perfectly matches the curve in option (4). The expression eKt dictates an exponential increase. Ensure the curve starts from a non-zero value (1) at t=0, as (N0/N0) = 1, rather than starting from the origin (0). Class 12 Chemistry: Chemical Kinetics

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 e-lambda t). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 e-lambda t). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging exponential profile starting from 1, which perfectly matches the curve in option (4). The expression eKt dictates an exponential increase. Ensure the curve starts from a non-zero value (1) at t=0, as (N0/N0) = 1, rather than starting from the origin (0). Class 12 Chemistry: Chemical Kinetics

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Exponential growth equation model: N = N0 eKt Normalized configuration formula: Radioactive decay follows a decreasing exponential path (N = N0 e-lambda t). Conversely, cell culture growth functions via an *increasing* exponential pattern because the rate of growth is directly proportional to the current population size (dN/dt = KN). This results in an exponential curve that starts at (N/N0) = 1 when t = 0 and curves sharply upward over time. Plotting (N/N0) against time shows an upward-clinging

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the…

Q30 jee_main_2025_03_april_morning First Order Kinetics

In a reaction

A+B ightarrow C

, initial concentrations of A and B are related as

[A]_0=8[B]_0

. The half lives of A and B are 10 min and 40 min. respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

A. 60 min
B. 80 min
C. 20 min
D. 40 min

Solution

### Related Formula For a first-order integrated rate law expression:

[A]_t = [A]_0 e^-k_A t quad textwhere k = fracln 2t_1/2

### Core Logic We require the instantaneous concentration to be identical at time

t

:

[A]_t = [B]_t implies [A]_0 e^-k_A t = [B]_0 e^-k_B t

frac[A]_0[B]_0 = e^(k_A - k_B)t

### Step 1: Substituting Parameters Substitute

[A]_0 = 8[B]_0

and express rate constants in terms of half-lives:

8 = e^(k_A - k_B)t implies ln 8 = (k_A - k_B)t

3ln 2 = ln 2 left( frac1(t_1/2)_A - frac1(t_1/2)_B ight) times t

3 = left( frac110 - frac140 ight) times t implies 3 = frac340 times t implies t = 40text min.

### Pattern Recognition Shortcut: Express the concentration drop using half-life indices:

[A]_t = frac[A]_02^t/10 = frac8[B]_02^t/10

[B]_t = frac[B]_02^t/40

Equating both:

8 cdot 2^-t/10 = 2^-t/40 implies 2^3 = 2^fract10 - fract40 implies 3 = frac3t40 implies t = 40text min.

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q jee_main_2025_04_april_evening Arrhenius Equation and Activation Energy

Consider the following plots of log of rate constant k (log k) vs

frac1mathrmT

for three different reactions. The correct order of activation energies of these reactions is

Arrhenius plots of log k vs 1 over T for Q40 - JEE Main 2025 Evening — The graph depicts three linear curves with distinct negative slopes showing the temperature dependence of rate constants.

A. $mathrmEa_2 > mathrmEa_1 > mathrmEa_3$
B. $mathrmEa_1 > mathrmEa_3 > mathrmEa_2$
C. $mathrmEa_1 > mathrmEa_2 > mathrmEa_3$
D. $mathrmEa_3 > mathrmEa_2 > mathrmEa_1$

Solution

### Related Formula

log k = log A - fracE_a2.303 R T

textSlope of the line = -fracE_a2.303 R implies |textSlope| propto E_a

### Core Logic From the given graph, we look at the steepness (magnitude of the negative slope) of lines 1, 2, and 3: - Line 2 is the steepest, meaning it has the largest slope magnitude. - Line 1 has an intermediate slope. - Line 3 is the flattest, indicating the smallest slope magnitude. Since the activation energy

E_a

is directly proportional to the magnitude of this slope:

|textSlope_2| > |textSlope_1| > |textSlope_3| implies E_a2 > E_a1 > E_a3

### Pattern Recognition In Arrhenius coordinates, steepness equals barriers. A steeper line means the reaction rate is highly sensitive to temperature because it has a higher activation energy (

E_a

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q45 jee_main_2025_04_april_evening Integrated Rate Equations

Half-life of zero order reaction

mathrmA rightarrow

product is 1 hour, when initial concentration of reaction is

2.0 mathrm~mol mathrmL^-1

. The time required to decrease concentration of A from 0.50 to

0.25 mathrm~mol mathrmL^-1

is:

A. 0.5 hour
B. 4 hour
C. 15 min
D. 60 min

Solution

### Related Formula

t_1/2 = frac[A]_02k quad text(for Zero-Order रिएक्शन)

t = frac[A]_0 - [A]_tk

### Core Logic 1. Find the rate constant

k

using the given half-life parameters:

1 text hour = 60 text min = frac2.02k implies k = frac2.02 times 60 = frac160 mathrm~M cdot min^-1

2. Calculate the time

t

to drop from

0.50 mathrm~molcdot L^-1

to

0.25 mathrm~molcdot L^-1

:

t = frac0.50 - 0.25k = frac0.25left(frac160right) = 0.25 times 60 = 15 text minutes

### Pattern Recognition For zero-order systems, the rate of reaction is entirely independent of concentration. This means the time required to consume a specific quantity of reactant scales linearly with the concentration change (

t = fracDelta Ck

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q32 jee_main_2025_04_april_morning Effect of Catalyst

For

A_2 + B_2 rightleftharpoons 2AB

,

E_a

for forward and backward reaction are

180

and

200mathrm~kJ~mol^-1

respectively. If catalyst lowers

E_a

for both reaction by

100mathrm~kJ~mol^-1

, which of the following statement is correct?

A. $textCatalyst does not alter the Gibbs energy change of a reaction.$
B. $textCatalyst can cause non-spontaneous reactions to occur.$
C. $textThe enthalpy change for the reaction is +20mathrm~kJ~mol^-1.$
D. $textThe enthalpy change for the catalysed reaction is different from that of uncatalysed reaction.$

Solution

### Related Formula

Delta H = E_a(f) - E_a(b)

### Core Logic A catalyst accelerates both forward and backward path steps symmetrically by carving a lower activation energy profile route. * Uncatalyzed values:

Delta H = 180 - 200 = -20mathrm~kJ~mol^-1

. * Catalyzed values:

E_a(f)' = 80mathrm~kJ~mol^-1

and

E_a(b)' = 100mathrm~kJ~mol^-1

, leading to

Delta H' = 80 - 100 = -20mathrm~kJ~mol^-1

. Thermodynamic parameters (

Delta H

,

Delta G

,

Delta S

) depend strictly on the initial and final energy states of reactants and products, meaning they are completely unaltered by the presence of a catalyst. ### Pattern Recognition Catalysts alter only kinetic properties (rate, activation barriers). They have zero impact on equilibrium positions or thermodynamic state parameters like

Delta G

or

Delta H

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

Q33 jee_main_2025_04_april_morning Rate Law and Order

Rate law for a reaction between A and B is given by

R = k[A]^n[B]^m

. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction

left(fracr_2r_1 ight)

is:

A. $2^(n-m)$ $2^{(n-m)}$
B. $(n-m)$ $(n-m)$
C. $(m+n)$ $(m+n)$
D. $frac12^m+n$ $\frac{1}{2^{m+n}}$

Solution

### Related Formula

r = k [A]^n [B]^m

### Core Logic Let the initial rate relation be:

r_1 = k [A]^n [B]^m

When concentration parameters shift (

[A]' = 2[A]

and

[B]' = frac[B]2

):

r_2 = k (2[A])^n left(frac[B]2right)^m = k cdot 2^n [A]^n cdot 2^-m [B]^m

r_2 = 2^(n-m) cdot left(k [A]^n [B]^mright) = 2^(n-m) cdot r_1

Taking the ratio yields:

fracr_2r_1 = 2^(n-m)

### Pattern Recognition Powers simplify cleanly via exponent rules: doubling a base scales the expression by

2^n

, while halving scales it by

2^-m

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics

For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Solution & Explanation

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 3

Solution

Solution

Solution

Solution

Solution

More Chemical Kinetics Questions — jee_main_2025_28_jan_evening

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For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth?

Solution & Explanation

Reference Study Guides

More Chemical Kinetics Previous-Year Questions — Page 3

Solution

Solution

Solution

Solution

Solution

More Chemical Kinetics Questions — jee_main_2025_28_jan_evening

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